• POJ 3255 Roadblocks(A*求次短路)


    Roadblocks
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 12167   Accepted: 4300

    Description

    Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

    The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

    The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

    Input

    Line 1: Two space-separated integers: N and R 
    Lines 2..R+1: Each line contains three space-separated integers: AB, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

    Output

    Line 1: The length of the second shortest path between node 1 and node N

    Sample Input

    4 4
    1 2 100
    2 4 200
    2 3 250
    3 4 100

    Sample Output

    450

    Hint

    Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

    Source

    题目链接:POJ 3255

    裸的A*,注意一点题目要强行次短路,不存在次短就来回跑一圈再跑到终点(MDZZ)……

    代码:

    #include<iostream>
    #include<algorithm>
    #include<cstdlib>
    #include<sstream>
    #include<cstring>
    #include<bitset>
    #include<cstdio>
    #include<string>
    #include<deque>
    #include<stack>
    #include<cmath>
    #include<queue>
    #include<set>
    #include<map>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define LC(x) (x<<1)
    #define RC(x) ((x<<1)+1)
    #define MID(x,y) ((x+y)>>1)
    typedef pair<int,int> pii;
    typedef long long LL;
    const double PI=acos(-1.0);
    const int N=100010;
    struct edge
    {
    	int to;
    	int pre;
    	int dx;
    };
    struct info
    {
    	int cur;
    	int g;
    	int h;
    	int f;
    	bool operator<(const info &b)const
    	{
    		return f>b.f;
    	}
    };
    info S;
    edge E[N<<1];
    int head[N<<1],ne;
    int d[N];
    priority_queue<info>q;
    void add(int s,int t,int d)
    {
    	E[ne].to=t;
    	E[ne].dx=d;
    	E[ne].pre=head[s];
    	head[s]=ne++;
    }
    void init()
    {
    	CLR(head,-1);
    	ne=0;
    	CLR(d,INF);
    	while (!q.empty())
    		q.pop();
    }
    void spfa(int s)
    {
    	priority_queue<pii>Q;
    	d[s]=0;
    	Q.push(pii(-d[s],s));
    	while (!Q.empty())
    	{
    		int now=Q.top().second;
    		Q.pop();
    		for (int i=head[now]; ~i; i=E[i].pre)
    		{
    			int v=E[i].to;
    			int w=E[i].dx;
    			if(d[v]>d[now]+w)
    			{
    				d[v]=d[now]+w;
    				Q.push(pii(-d[v],v));
    			}
    		}
    	}
    }
    int main(void)
    {
    	int n,r,i,a,b,c;
    	while (~scanf("%d%d",&n,&r))
    	{
    		init();
    		for (i=0; i<r; ++i)
    		{
    			scanf("%d%d%d",&a,&b,&c);
    			add(a,b,c);
    			add(b,a,c);
    		}
    		spfa(n);
    		S.g=0;
    		S.h=d[1];
    		S.cur=1;
    		S.f=S.g+S.h;
    		int second_dx=d[1];
    		q.push(S);
    		while (!q.empty())
    		{
    			info now=q.top();
    			q.pop();
    			if(now.cur==n)
    			{
    				if(second_dx!=now.f)
    				{
    					second_dx=now.f;
    					break;
    				}
    			}
    			for (i=head[now.cur]; ~i; i=E[i].pre)
    			{
    				info v;
    				v.cur=E[i].to;
    				v.g=now.g+E[i].dx;
    				v.h=d[v.cur];
    				v.f=v.g+v.h;
    				q.push(v);
    			}
    		}
    		printf("%d
    ",second_dx);
    	}
    	return 0;
    }
  • 相关阅读:
    食物链(并查集)
    字母奔跑
    19+199+1999+……+1999…9(1999个9)和是多少?
    利用union判断CPU是大端模式还是小端模式
    闰年个数(非循环)
    计算catlan数f(n)(动态规划)
    模拟循环调度(队列)
    回溯表达式,+不小于4个,乘号不小于2个,列出表达式种数
    枚举{1,2,3,4}子集
    回溯n个元素的子集
  • 原文地址:https://www.cnblogs.com/Blackops/p/5766262.html
Copyright © 2020-2023  润新知